《2025年启东中学作业本九年级数学上册苏科版连淮专版》

8. 如图,在$5×3$的网格图中,每个小正方形的边长均为1,设经过图中格点$A,C,B的圆弧与AE交于点H$,则$\overset{\frown }{AH}$的长为 (B)
A.$\frac {\sqrt {13}}{6}π$
B.$\frac {\sqrt {13}}{4}π$
C.$\frac {\sqrt {5}}{3}π$
D.$\frac {\sqrt {5}}{2}π$

答案： B

9. 如图,在$\triangle ABC$中,$AB= AC$,点$O在边AC$上,以点$O$为圆心,4为半径的圆恰好过点$C$,且与边$AB相切于点D$,交$BC于点E$,则劣弧$DE$的长是$2\pi$.(结果保留$π$)

答案： $2\pi$

10. (2024·资阳)如图,在矩形$ABCD$中,$AB= 4,AD= 2$.以点$A$为圆心,$AD长为半径作弧交AB于点E$,再以$AB$为直径作半圆,与$\overset{\frown }{DE}交于点F$,则图中阴影部分的面积为$\sqrt{3} + \frac{2}{3}\pi$.

答案： $\sqrt{3} + \frac{2}{3}\pi$

11. 如图,$AB为\odot O$的直径,点$C在\odot O$上,延长$BC至点D$,使得$DC= CB$,连接$DA$并延长,与$\odot O交于点E$,连接$AC,CE$.
(1)求证:$∠D= ∠E$;
(2)若$AB= 4,\overset{\frown }{AC}的长度为\frac {2}{3}π$,求阴影部分的面积.

答案：

(1) 证明: $\because AB$ 是 $\odot O$ 的直径, $\therefore AC \perp BD$.
又 $\because DC = BC$, $\therefore AC$ 垂直平分 $BD$,
$\therefore AD = AB$, $\therefore \angle D = \angle B$.
$\because \angle B = \angle E$, $\therefore \angle D = \angle E$.
(2) 解: 如答图, 连接 $OC$, 过点 $O$ 作 $OF \perp BC$ 于点 $F$.
$\because AB = 4$, $\therefore OB = OA = 2$.
设 $\angle AOC = n^{\circ}$, 由弧长公式得 $\frac{2n\pi}{180} = \frac{2\pi}{3}$,
$\therefore n = 60$, 即 $\angle AOC = 60^{\circ}$.
$\because OB = OC$, $\therefore \angle OBC = \angle OCB$, 而 $\angle AOC = \angle OBC + \angle OCB$,
$\therefore \angle B = 30^{\circ}$, $AC = \frac{1}{2}AB = 2$, $OF = \frac{1}{2}OB = 1$, 由勾股定理, 得 $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - 2^2} = 2\sqrt{3}$,
$\therefore S_{阴影} = S_{扇形OAC} + S_{\triangle BOC} = \frac{60}{360}\pi × 2^2 + \frac{1}{2} × 2\sqrt{3} × 1 = \frac{2\pi}{3} + \sqrt{3}$.

12. 如图,在$\triangle ABC$中,$AB= AC$,以$AB为直径的\odot O分别与BC,AC交于点D,E$,过点$D作DF⊥AC于点F$,连接$AD,DE$.
(1)若$\odot O的半径为3,∠CDF= 15^{\circ }$,求阴影部分的面积;
(2)求证:$DF是\odot O$的切线;
(3)求证:$∠EDF= ∠DAC$.

答案：

(1) 解: 如答图, 过点 $O$ 作 $OM \perp AE$ 于点 $M$, 连接 $OE$.
$\because AB$ 是 $\odot O$ 的直径, $\therefore AD \perp BC$, $\therefore \angle DAC + \angle C = 90^{\circ}$.
$\because AB = AC$, $\therefore DB = DC$, $\angle DAB = \angle DAC$.
$\because DF \perp AC$, $\therefore \angle CDF + \angle C = 90^{\circ}$,
$\therefore \angle DAC = \angle CDF = 15^{\circ}$, $\therefore \angle OAE = 30^{\circ}$.
$\because OA = OE$, $\therefore \angle AOE = 120^{\circ}$.
$\because OM \perp AE$, $\therefore OM = \frac{1}{2}OA = \frac{3}{2}$, $AM = \frac{3}{2}\sqrt{3}$,
$\therefore AE = 2AM = 3\sqrt{3}$,
$\therefore S_{阴影} = S_{扇形OAE} - S_{\triangle OAE} = \frac{120\pi × 3^2}{360} - \frac{1}{2} × 3\sqrt{3} × \frac{3}{2} = 3\pi - \frac{9\sqrt{3}}{4}$.

(2) 证明: 如答图, 连接 $OD$.
$\because OA = OB$, $DB = DC$, $\therefore OD // AC$.
$\because DF \perp AC$, $\therefore \angle DFC = 90^{\circ}$,
$\therefore \angle ODF = \angle DFC = 90^{\circ}$, $\therefore DF \perp OD$.
$\because OD$ 为 $\odot O$ 的半径, $\therefore DF$ 是 $\odot O$ 的切线.
(3) 证明: $\because \angle DAB = \angle DAC$, $\therefore \overset{\frown}{DB} = \overset{\frown}{DE}$.
$\because DB = DC$, $\therefore DE = DC$.
$\because DF \perp AC$, $\therefore \angle EDF = \angle CDF$.
$\because \angle CDF = \angle DAC$, $\therefore \angle EDF = \angle DAC$.